Problem: Let $f(x) = 8x^{2}-4x-9$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $8x^{2}-4x-9 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 8, b = -4, c = -9$ $ x = \dfrac{+ 4 \pm \sqrt{(-4)^{2} - 4 \cdot 8 \cdot -9}}{2 \cdot 8}$ $ x = \dfrac{4 \pm \sqrt{304}}{16}$ $ x = \dfrac{4 \pm 4\sqrt{19}}{16}$ $x =\dfrac{1 \pm \sqrt{19}}{4}$